8.51 The enthalpy of formation of trinitrotoluene (TNT)

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Lisa Nguyen 1D
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8.51 The enthalpy of formation of trinitrotoluene (TNT)

Postby Lisa Nguyen 1D » Sat Jan 16, 2016 5:39 pm

The enthalpy of formation of trinitrotoluene (TNT) is -67 kJ·mol^-1, and the density of TNT is 1.65 g·cm^-3. In principle, it could be used as a rocket fuel, with the gases resulting from its decomposition streaming out of the rocket to give the required thrust.
In practice, of course, it would be extremely dangerous as a fuel because it is sensitive to shock. Explore its potential as a rocket fuel by calculating its enthalpy density (enthalpy released per liter) for the reaction 4 C7H5N3O6 (s) +  21 O2 (g) --> 28 CO2 (g) + 10 H2O (g) + 6 N2 (g).

Could someone explain to me how to solve this problem?

In the solution, it says that the enthalpy of reaction may be found using the enthalpies of formation: 28 (-393.51 KJ·mol^-1) +10 (-241.82 KJ·mol^-1) - 4(-67 KJ·mol^-1) = -13168 KJ·mol^-1

1) What happened to the 21 O2(g) and the 6 N2 (g)? How come they aren't included in the enthalpy of reaction? Is it because they are in their standard state?
2) Why is 28 mol CO2 and 10 mol H20 added while 4 mole of TNT is subtracted?

1/4 of this amount of energy (3292 KJ·mol^-1) will be released per mole of TNT consumed. The energy density in KJ per mole may be found by dividing this amount of energy with the mass of one mole of TNT and then by multiplying with the density of TNT:
3292 KJ·mol^-1/ 227.14 g·mol^-1(1.65 g·cm^-3)(10^3cm^3·L^-1) = +23.9 x 10^3 KJ·L^-1

3) How do I know that 1/4 of this energy will be released per mole of TNT consumed?
4) Why is this energy (3292 KJ·mol^-1) taken and divided with the mass of one mole of TNT instead of -13168 KJ·mol^-1?

Jake Ney lecture 1 discussion 1F
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Joined: Fri Sep 25, 2015 3:00 am

Re: 8.51 The enthalpy of formation of trinitrotoluene (TNT)

Postby Jake Ney lecture 1 discussion 1F » Sat Jan 16, 2016 6:09 pm

That is correct the O2 and N2 are diatomic gases in their most stable form and thus their standard enthalpy of formation is zero. Using the enthalpies of formation we calculate that CO2 requires -393.51 kJ/mol while H20 requires -241.82. Given TNT's enthalpy of formation we calculate:

28(-393.51 kJ/mol)+10(-241.82)-4(-67)=-13168 kJ/mol.

TNT is negative while CO2 and H2O are positive because the enthalpy of the reaction is equal to the sum of the products minus the sum of the reactants.

-13168 kJ/mol is the value released for the total reaction so we divide this value by 4 getting 3292 kJ/mol before converting this value to kJ/L using the molar mass of TNT as well as its density.

Leila_4G
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Joined: Sat Sep 14, 2019 12:17 am

Re: 8.51 The enthalpy of formation of trinitrotoluene (TNT)

Postby Leila_4G » Tue Feb 11, 2020 11:10 am

Thank you! This helped so much!


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